1985 AHSME Problems/Problem 29

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Problem

In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$?

$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851$

Solution

By the formula for the sum of a geometric series, \begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*} and similarly $$b = \frac{5\left(10^{1985}-1\right)}{9},$$ so \begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}

We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$, with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$, with $1$ two and $1986$ zeroes. Subtracting therefore gives $$10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,$$ where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$, now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.

Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$, it follows that $$\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,$$ with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$, $8 \cdot 4 = 32$, and $8 \cdot 4 + 3 = 35$, we obtain $$\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,$$ where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{\text{(C)} \ 17865}.\end{align*}